Problem: A few families took a trip to an amusement park together. Tickets cost $$7.50$ each for adults and $$2.50$ each for kids, and the group paid $$27.50$ in total. There were $3$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Solution: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7.5x+2.5y = 27.5}$ ${x = y-3}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-3}$ for $x$ in the first equation. ${7.5}{(y-3)}{+ 2.5y = 27.5}$ Simplify and solve for $y$ $ 7.5y-22.5 + 2.5y = 27.5 $ $ 10y-22.5 = 27.5 $ $ 10y = 50 $ $ y = \dfrac{50}{10} $ ${y = 5}$ Now that you know ${y = 5}$ , plug it back into ${x = y-3}$ to find $x$ ${x = }{(5)}{ - 3}$ ${x = 2}$ You can also plug ${y = 5}$ into ${7.5x+2.5y = 27.5}$ and get the same answer for $x$ ${7.5x + 2.5}{(5)}{= 27.5}$ ${x = 2}$ There were $2$ adults and $5$ kids.